Famous Geometry Theorems - Department of Mathematics - The ...

Mathematical Excalibur, Vol. 10, No. 3, Aug. 05- Sept. 05 Page 4
a 4 ≤ n − 1. By condition (2), we have
1+a 2 +a 3 +a 4 =2n. (A)
For k = 1, 2, …, n − 1, let
f i (k) = [ka i /n] − [(k − 1)a i /n],
then f i (k) ≤ (ka i /n) + 1 − (k − 1)a i /n = 1
+ (a i /n) < 2. So f i (k) = 0 or 1. Since x =
[x/n]n + (x) n , subtracting the case x =
ka i from the case x = (k − 1)a i , then
summing i = 1, 2, 3, 4, using condition
(2) and (A), we get
f 1 (k) + f 2 (k) + f 3 (k) + f 4 (k) = 2.
Since a 1 = 1, we see f 1 (k)=0 and exactly
two of f 2 (k), f 3 (k), f 4 (k) equal 1. (B)
Since a i 2, by
(C), we get f 2 (s + 1) = 0 and f 2 (t 4 + 1) =
0. So s + 1 ≠ t 4 , which implies f 2 (t 4 − 1)
= 0 by the definition of s. Then k = t 4 −
1 or t 4 + 1 contradicts (B).
So t 4 ≥ n, then n − a 4 = 1. We get a 1 + a 4
= n = a 2 + a 3 .
Problem 230. Let k be a positive
integer. On the two sides of a river,
there are in total at least 3 cities. From
each of these cities, there are exactly k
routes, each connecting the city to a
distinct city on the other side of the river.
Via these routes, people in every city can
reach any one of the other cities.
Prove that if any one route is removed,
people in every city can still reach any one
of the other cities via the remaining
routes.
(Source: 1996 Iranian Math Olympiad,
Round 2)
Solution. LEE Kai Seng (HKUST).
Associate each city with a vertex of a
graph. Suppose there are X and Y cities to
the left and to the right of the river
respectively. Then the number of routes
(or edges of the graph) in the beginning is
Xk = Yk so that X = Y. We have X + Y ≥ 3.
After one route between city A and city B
is removed, assume the cities can no
longer be connected via the remaining
routes. Then each of the other cities can
only be connected to exactly one of A or B.
Then the original graph decomposes into
two connected graphs G A and G B , where
G A has A as vertex and G B has B as vertex .
Let X A be the number of cities among the X
cities on the left sides of the river that can
still be connected to A after the route
between A and B was removed and
similarly for X B , Y A , Y B . Then the number
of edges in G A is X A k-1 = Y A k. Then (X A -
Y A )k = 1. So k = 1. Then in the beginning
X = 1 and Y = 1, contradicting X + Y ≥ 3.
Olympiad Corner
(continued from page 1)
Problem 4. Consider the sequence a 1 ,
a 2 , … defined by
a n = 2 n + 3 n + 6 n − 1 (n = 1, 2, …)
Determine all positive integers that are
relatively prime to every term of the
sequence.
Problem 5. Let ABCD be a given convex
quadrilateral with sides BC and AD equal
in length and not parallel. Let E and F be
interior points of the sides BC and AD
respectively such that BE = DF. The lines
AC and BD meet at P, the lines BD and EF
meet at Q, the lines EF and AC meet at R.
Consider all the triangles PQR as E and F
vary. Show that the circumcircles of these
triangles have a common point other than
P.
Problem 6. In a mathematical
competition 6 problems were posed to
the contestants. Each pair of problems
was solved by more than 2/5 of the
contestants. Nobody solved all 6
problems. Show that there were at
least 2 contestants who each solved
exactly 5 problems.
Famous Geometry
Theorems
(continued from page 2)
Example 5 (2005 Chinese Math
Olympiad) A circle meets the three
sides BC, CA, AB of triangle ABC at
points D 1 , D 2 ; E 1 , E 2 and F 1 , F 2 in turn.
The line segments D 1 E 1 and D 2 F 2
intersect at point L, line segments E 1 F 1
and E 2 D 2 intersect at point M, line
segments F 1 D 1 and F 2 E 2 intersect at
point N. Prove that the three lines AL,
BM and CN are concurrent.
B
P
A
F 1 E 2
F 2
D 1
D 2
E 1
Solution. Let P = D 1 F 1 ∩ D 2 E 2 , Q =
E 1 D 1 ∩ E 2 F 2 , R = F 1 E 1 ∩ F 2 D 2 .
Applying Pascal’s Theorem to E 2 , E 1 ,
D 1 , F 1 , F 2 , D 2 , we get A, L, P are
collinear. Applying Pascal’s Theorem
to F 2 , F 1 , E 1 , D 1 , D 2 , E 2 , we get B, M, Q
are collinear. Applying Pascal’s
Theorem to D 2 , D 1 , F 1 , E 1 , E 2 , F 2 , we
get C, N, R are collinear.
Let X = E 2 E 1 ∩ D 1 F 2 = CA ∩ D 1 F 2 , Y =
F 2 F 1 ∩ E 1 D 2 = AB ∩ E 1 D 2 , Z = D 2 D 1 ∩
F 1 E 2 = BC ∩ F 1 E 2 . Applying Pascal’s
Theorem to D 1 , F 1 , E 1 , E 2 , D 2 , F 2 , we
get P, R, X are collinear. Applying
Pascal’s Theorem to E 1 , D 1 , F 1 , F 2 , E 2 ,
D 2 , we get Q, P, Y are collinear.
Applying Pascal’s Theorem to F 1 , E 1 ,
D 1 , D 2 , F 2 , E 2 , we get R, Q, Z are
collinear.
For △ABC and △PQR, we have X =
CA ∩ RP, Y = AB ∩ PQ, Z = BC ∩ QR.
By the converse of Desargues’
Theorem, lines AP = AL, BQ = BM,
CR = CN are concurrent.
L
C